My homework is:
Find a nontrivial continuous action $\mathbb{Z}_n \to S^1$ of the cyclic group of order n on the circle and compute the quotient space?
My idea is:
action could be: $n*p = (-1)^n * p$. It would mean that all points lying on one line, but there are just two point lying on one line, because it is $S^1$, would be identical and the quotient would be space $\mathbb{RP}^1$.
I would like to know if I understand action in a good way. About me is $n*p$ a symbol for an action. Let $p_1, p_2 \in S^n$ and quotient will be the metric space, where $p_1 \sim p_2$ iff and only iff $p_1 = p_2$ or $p_1 = -p_2$. Would it be the way how to create a quotient on $S^1$ under a relation equivalence, which identify the points $p$ and $-p$? And it will give me a topological space $\mathbb{RP}^1$. Do you think that it is a correct explenation? Thanks :)
Note that your action is trivial when $n$ is even. It does not even make much sense, since the number $n$ is not an element of $\mathbb{Z}_n$.
Instead, consider the following. Let $S^1$ be identified with the unit circle $S^1 \simeq \{e^{i\varphi}: \varphi \in \mathbb{R}\}$ on the complex plane and let $\mathbb{Z}_n$ be identified with the group of roots of unity, that is $\mathbb{Z}_n \simeq \{e^{2i\pi k/n} : k = 0,1,\dotsc,n-1\}$. Now, let $\mathbb{Z}_n$ act on $S^1$ by the formula $$ (e^{2i\pi k/n},e^{i\varphi}) \longmapsto e^{2i\pi k/n}\,e^{i\varphi} = e^{i(\varphi+2\pi k/n)}. $$ This action is nontrivial for any $n>0$. It is continuous since the topology on $\mathbb{Z}_n$ is discrete. The corresponding quotient map $S^1 \to S^1/\mathbb{Z_n}$ is actually the $n$-fold covering map.
In particular, when $n=2$ and $\mathbb{Z}_2$ is of the form $\mathbb{Z}_2 = \{1,-1\}$, our action reads $$ (\pm 1,e^{i\varphi}) \longmapsto \pm e^{i\varphi}. $$ Here, the quotient space is $S^1/(z \sim -z)$, and this is, of course, the projective line $\mathbb{R}P^1$.